How do you solve ${log}_{a} x - {log}_{a} 2 = {log}_{a} 9 + {log}_{a} 8$ $Log_a x - Log_a 2 = Log_a 9 + Log_a 8$ ?

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How do you solve ${log}_{a} x - {log}_{a} 2 = {log}_{a} 9 + {log}_{a} 8$ $Log_a x - Log_a 2 = Log_a 9 + Log_a 8$ ?

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Answer 1 · 2016-03-06T15:32:08

$x = 144$ $x=144$

Explanation:

${log}_{a} (\frac{x}{2}) = {log}_{a} 3^{2} + {log}_{a} 2^{3}$ $log_a( x/2)=log_a3^2+log_a2^3$
${log}_{a} (\frac{x}{2}) = {log}_{a} (3^{2} \cdot 2^{3})$ $log_a( x/2)=log_a(3^2*2^3)$
$\frac{x}{2} = 3^{2} \cdot 2^{3}$ $x/2=3^2*2^3$
$x = 3^{2} \cdot 2^{4}$ $x=3^2*2^4$
$x = 9 \cdot 16$ $x=9*16$
$x = 144$ $x=144$

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How do you solve ${log}_{a} x - {log}_{a} 2 = {log}_{a} 9 + {log}_{a} 8$ $Log_a x - Log_a 2 = Log_a 9 + Log_a 8$ ?

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How do you solve ${log}_{a} x - {log}_{a} 2 = {log}_{a} 9 + {log}_{a} 8$ $Log_a x - Log_a 2 = Log_a 9 + Log_a 8$ ?